Assume the wave equation in two dimensions: $$ \begin{cases} u_{xx}+u_{yy} = u_{tt}\\ u(x,y,t=0) = f(x,y) \\ u_t(x,y,t=0) = g(x,y) \end{cases} $$
where $x$ and $y$ represent spatial variables (Cartesian Coordinates) and $t$ represents time. Subscripts represent differentiation w-r-t the corresponding variable.
This system of equation can be solved as well in the following representation:
$ \begin{cases} v_y+w_x = q_t \\ w_t = q_x \\ v_t = q_y \end{cases} $
where I have the "freedom" to choose initial conditions:
$ \begin{cases} v(x,y,t=0) = f^a(x,y) \\ w(x,y,t=0) = f^b(x,y) \\ q(x,y,t=0) = g(x,y) \end{cases}$
My question: let's assume I am choosing $f^a = \cos(xy)$ and $f^b = 0$. These functions do not represent any valid initial condition in the original wave equation. I.E. to find $f$ relating $f^a$ and $f^b$, I must find a function such that:
$ \bigg(-\sin(xy)/x\bigg)_x+\tilde{f}_x(x) = 0$
; which cannot exist. Thus, there is no $f$ possible and I, in fact, do not solve the wave equation.
Hence, is there a notion which relates the wave equation and the second and first representations in this case? (maybe a weak solution but I am not proficient in this so I am just guessing)
If no, then to solve the wave equations, must I have $f^a = f_x$ and $f^b = f_y$? (which works in a piecewise continuous/ weak sense)
Edit: As per Willie Wong's comment, I do need an integrability condition (IntC):
$w_y = v_x$
in order to relate my presentation to the wave equation.
However, the IntC is generally omitted in the representation used in finite volume methods when solving for the wave equation. Therefore, I believe there must be some reason for this representation to be used.